December 23rd, 2012 at 9:01:02 AM
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Having been a regular member of several other roulette forums, I have decided to bring my thoughts to this forum, as this seems to be the best place of discussion for the math and odds of betting matters and I am looking for feedback, specifically how best to apply this to real world betting.

First I need to clarify that when I speak of a bet, that is not the same as a spin. A bet can consist of a single spin, or any number of spins. A bet is playing however many spins come to the ultimate resolution of a "win" or "loss", which I will explain now.

This system has nothing to do with bet selection, nor is it really a money management system. What it is is a method of betting that holds a distinct mathematical advantage over the house, even on a 00 table.

For the sake of simplicity, I will give the following example, but it doesn't matter how you choose your dozens.

If the 2nd dozen has hit, you have two options for movement - Either the spin can move to the 1st dozen (left) or the 3rd dozen (right). A bet requires however many spins it takes to move to the left or right. If I get a series of 2-2-2-2-2-2-2-2-1, that is 9 spins but only one bet.

In this example, I anticipate a move to the right, meaning I will bet with the expectation that the spin will eventually move from the 2nd dozen to the 3rd dozen. If it moves to the right (3rd dozen), it is a "win" and the bet is over. If it moves to the left (1st dozen), it is a "loss" and the bet is over. For as long as it repeats on the 2nd dozen, the bet is not complete and we continue spinning until the bet is resolved.

I will place 1 unit on the 2nd dozen and 1 unit on the 3rd dozen. Every time it lands on either, I have +1 and if it lands on 1st dozen, I am -2 units.

Here is where the math moves in our favor: The most I can possibly lose on this bet is -2 units, whereas I have unlimited winning potential for as long as it repeats on the 2nd dozen. I will be explaining the math using a single 0 table.

On the first spin, we have a 13/37 chance of losing (1st dozen + 0), a 12/37 chance of winning (3rd dozen), and a 12/37 chance of repeating and moving to the second spin of the bet. Our expected value of this first spin is -2.7%. However, 32.4% of the time, we repeat our previous dozen and move to the second spin of the bet. As we have already gained +1 unit from the repeat on the 2nd dozen, our new payout is (0+1st dozen)=-1 [-2 units offset by our previous win], 3rd dozen=+2 (win +1 plus the +1 from previous repeat), and 2nd dozen=+2 repeats and move on to the third spin of the bet.

Third spin (which happens 10.5% of the time (12/37)^2) has payouts of (0+1st dozen)=0, 3rd dozen=+3 and 2nd dozen=+3 and move to the fourth spin, etc.

In 37 spins, we would average 13 losses of -2, 12 wins of +1, and 12 spins where we move to higher levels. Of these 12, about 4 would be losses of -1, 4 would be wins of +2, and 4 would move on again. Of these, 1 would be a loss of -0, 1 would be a win of +3, and 1 or 2 would continue on to higher winnings.

By extending the odds of each level out, this method gives a positive expectation of +3.xxx units over a single 0 table, better than +1.xxx on a 0/00 table, and better than +5.xxx on no-zero tables (if you are lucky enough to find them).

First I need to clarify that when I speak of a bet, that is not the same as a spin. A bet can consist of a single spin, or any number of spins. A bet is playing however many spins come to the ultimate resolution of a "win" or "loss", which I will explain now.

This system has nothing to do with bet selection, nor is it really a money management system. What it is is a method of betting that holds a distinct mathematical advantage over the house, even on a 00 table.

For the sake of simplicity, I will give the following example, but it doesn't matter how you choose your dozens.

If the 2nd dozen has hit, you have two options for movement - Either the spin can move to the 1st dozen (left) or the 3rd dozen (right). A bet requires however many spins it takes to move to the left or right. If I get a series of 2-2-2-2-2-2-2-2-1, that is 9 spins but only one bet.

In this example, I anticipate a move to the right, meaning I will bet with the expectation that the spin will eventually move from the 2nd dozen to the 3rd dozen. If it moves to the right (3rd dozen), it is a "win" and the bet is over. If it moves to the left (1st dozen), it is a "loss" and the bet is over. For as long as it repeats on the 2nd dozen, the bet is not complete and we continue spinning until the bet is resolved.

I will place 1 unit on the 2nd dozen and 1 unit on the 3rd dozen. Every time it lands on either, I have +1 and if it lands on 1st dozen, I am -2 units.

Here is where the math moves in our favor: The most I can possibly lose on this bet is -2 units, whereas I have unlimited winning potential for as long as it repeats on the 2nd dozen. I will be explaining the math using a single 0 table.

On the first spin, we have a 13/37 chance of losing (1st dozen + 0), a 12/37 chance of winning (3rd dozen), and a 12/37 chance of repeating and moving to the second spin of the bet. Our expected value of this first spin is -2.7%. However, 32.4% of the time, we repeat our previous dozen and move to the second spin of the bet. As we have already gained +1 unit from the repeat on the 2nd dozen, our new payout is (0+1st dozen)=-1 [-2 units offset by our previous win], 3rd dozen=+2 (win +1 plus the +1 from previous repeat), and 2nd dozen=+2 repeats and move on to the third spin of the bet.

Third spin (which happens 10.5% of the time (12/37)^2) has payouts of (0+1st dozen)=0, 3rd dozen=+3 and 2nd dozen=+3 and move to the fourth spin, etc.

In 37 spins, we would average 13 losses of -2, 12 wins of +1, and 12 spins where we move to higher levels. Of these 12, about 4 would be losses of -1, 4 would be wins of +2, and 4 would move on again. Of these, 1 would be a loss of -0, 1 would be a win of +3, and 1 or 2 would continue on to higher winnings.

By extending the odds of each level out, this method gives a positive expectation of +3.xxx units over a single 0 table, better than +1.xxx on a 0/00 table, and better than +5.xxx on no-zero tables (if you are lucky enough to find them).

December 23rd, 2012 at 9:08:05 AM
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the expected value of every spin is -2.7%. If you're showing anything else on any spin, you have a math error somewhere...

"So as the clock ticked and the day passed, opportunity met preparation, and luck happened." - Maurice Clarett

December 23rd, 2012 at 9:09:16 AM
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deleted

DUHHIIIIIIIII HEARD THAT!

December 23rd, 2012 at 9:09:38 AM
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deleted.

DUHHIIIIIIIII HEARD THAT!

December 23rd, 2012 at 9:19:43 AM
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I agree that the expected value of a spin is -2.7%. If I were to bet statically on 2nd and 3rd dozens, I would lose 2.7% of my stake every spin. However, the expected value of a "bet" is greater than this because it is only extended by winning spins. Every time you get a repeat within a "bet", the payoff changes. in the roughly 1-in-9 times that you get 2 repeats, you will be to a new payout table of 0/3/3 instead of -2/1/1.

As soon as a "bet" has come to an end, with either a win or a loss, you take your money and start a new "bet", for as many spins as that takes.

As soon as a "bet" has come to an end, with either a win or a loss, you take your money and start a new "bet", for as many spins as that takes.

December 23rd, 2012 at 9:22:20 AM
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I also need to clarify that the new "bet" is based on a new starting position, either the 1st dozen after a loss or the 3rd dozen after a win. That dozen would be the new repeat dozen where we would continue betting while we wait for a new win or loss on either of the other two dozens, however we choose to predict where the next spin will fall.

December 23rd, 2012 at 9:22:43 AM
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Your logic is flawed. The House Edge on all your money is 2.7%, no matter what.

Messing with the math doesn't change it, it only makes you more likely to make an error, which seems like what happened here. The Derivation of this edge comes from the fact that every bet pays of as if there are 36 numbers, but there are 37 or 38.

Messing with the math doesn't change it, it only makes you more likely to make an error, which seems like what happened here. The Derivation of this edge comes from the fact that every bet pays of as if there are 36 numbers, but there are 37 or 38.

December 23rd, 2012 at 9:26:37 AM
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No, it is only -2.7 on the first spin. Once you have already taken the +1 from a repeat, your new payouts would be 13x-1, and 24x+2, meaning that once you have gotten a repeat, your spins expected value is +0.3. It is critical to understand that a "bet" is different than a "spin" and the bet is not on a stagnate set of numbers. It changes according to your own play style, none of which I condemn or condone. This is all about how we win on a win, lose on a loss, and pocket a bunch of money from the repeats.

December 23rd, 2012 at 9:27:51 AM
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Quote:rdw4potusthe expected value of every spin is -2.7%. If you're showing anything else on any spin, you have a math error somewhere...

HA! you fell into the trap of not recognizing that:

"when I speak of a bet, that is not the same as a spin"

the next time Dame Fortune toys with your heart, your soul and your wallet, raise your glass and praise her thus: “Thanks for nothing, you cold-hearted, evil, damnable, nefarious, low-life, malicious monster from Hell!” She is, after all, stone deaf. ... Arnold Snyder

December 23rd, 2012 at 9:30:15 AM
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Why this is confusing at the beginning is that we are accustomed to only seeing a loss as -2 and a win as only +1. With the power of the repeats behind us, our losses could be -2, but they could also be +3 if they came after a series of repeats. Our wins could be +1, or they could be +6 after repeats. The bet is not over until we have a firm win or loss.